Load the student scores for the test - here we load the 2018 ETH Zurich test data:
head(test_scores)
## # A tibble: 6 x 38
## year class A1 A2 A3 A4 A5 A6 A7 A8 A9 A10 A11
## <chr> <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 2018 401-0~ 1 0 1 1 1 1 1 0 0 1 1
## 2 2018 401-0~ 1 0 1 1 0 0 1 1 1 0 0
## 3 2018 401-0~ 1 1 1 1 1 1 1 0 1 1 1
## 4 2018 401-0~ 1 0 0 0 0 0 0 0 0 0 0
## 5 2018 401-0~ 0 0 1 0 0 0 0 0 1 1 0
## 6 2018 401-0~ 1 1 1 1 0 1 1 1 0 0 0
## # ... with 25 more variables: A12 <dbl>, A13 <dbl>, A14 <dbl>, A15 <dbl>,
## # A16 <dbl>, A17 <dbl>, A18 <dbl>, A19 <dbl>, A20 <dbl>, A21 <dbl>,
## # A22 <dbl>, A23 <dbl>, A24 <dbl>, A25 <dbl>, A26 <dbl>, A27 <dbl>,
## # A28 <dbl>, A29 <dbl>, A30 <dbl>, A31 <dbl>, A32 <dbl>, A33 <dbl>,
## # A34 <dbl>, A35 <dbl>, A36 <dbl>
The number of responses from each class:
test_scores %>%
group_by(year, class) %>%
tally() %>%
gt() %>%
data_color(
columns = c("n"),
colors = scales::col_numeric(palette = c("Blues"), domain = NULL)
)
## Warning: The `.dots` argument of `group_by()` is deprecated as of dplyr 1.0.0.
| class | n |
|---|---|
| 2018 | |
| 401-0131-00L | 230 |
| 401-0141-00L | 124 |
| 401-0151-00L | 201 |
| 401-0231-10L | 170 |
| 401-0251-00L | 163 |
| 401-0261-G0L | 283 |
| 401-0261-GUL | 31 |
| 401-0261-GXL | 314 |
| 401-0271-00L | 108 |
| 401-0281-00L | 59 |
| 401-0291-00L | 235 |
| 401-1151-00L | 275 |
Mean and standard deviation for each item:
test_scores %>%
select(-class) %>%
group_by(year) %>%
skim_without_charts() %>%
select(-contains("character."), -contains("numeric.p"), -skim_type) %>%
group_by(year) %>%
gt() %>%
fmt_number(columns = contains("numeric"), decimals = 3) %>%
data_color(
columns = c("numeric.mean"),
colors = scales::col_numeric(palette = c("Greens"), domain = NULL)
) %>%
cols_label(
numeric.mean = "Mean",
numeric.sd = "SD"
)
| skim_variable | n_missing | complete_rate | Mean | SD |
|---|---|---|---|---|
| 2018 | ||||
| A1 | 0 | 1 | 0.964 | 0.186 |
| A2 | 0 | 1 | 0.293 | 0.455 |
| A3 | 0 | 1 | 0.654 | 0.476 |
| A4 | 0 | 1 | 0.644 | 0.479 |
| A5 | 0 | 1 | 0.502 | 0.500 |
| A6 | 0 | 1 | 0.745 | 0.436 |
| A7 | 0 | 1 | 0.731 | 0.444 |
| A8 | 0 | 1 | 0.519 | 0.500 |
| A9 | 0 | 1 | 0.488 | 0.500 |
| A10 | 0 | 1 | 0.556 | 0.497 |
| A11 | 0 | 1 | 0.610 | 0.488 |
| A12 | 0 | 1 | 0.728 | 0.445 |
| A13 | 0 | 1 | 0.377 | 0.485 |
| A14 | 0 | 1 | 0.586 | 0.493 |
| A15 | 0 | 1 | 0.485 | 0.500 |
| A16 | 0 | 1 | 0.205 | 0.404 |
| A17 | 0 | 1 | 0.632 | 0.482 |
| A18 | 0 | 1 | 0.368 | 0.482 |
| A19 | 0 | 1 | 0.343 | 0.475 |
| A20 | 0 | 1 | 0.742 | 0.437 |
| A21 | 0 | 1 | 0.536 | 0.499 |
| A22 | 0 | 1 | 0.541 | 0.498 |
| A23 | 0 | 1 | 0.463 | 0.499 |
| A24 | 0 | 1 | 0.610 | 0.488 |
| A25 | 0 | 1 | 0.748 | 0.434 |
| A26 | 0 | 1 | 0.834 | 0.372 |
| A27 | 0 | 1 | 0.422 | 0.494 |
| A28 | 0 | 1 | 0.481 | 0.500 |
| A29 | 0 | 1 | 0.534 | 0.499 |
| A30 | 0 | 1 | 0.749 | 0.434 |
| A31 | 0 | 1 | 0.394 | 0.489 |
| A32 | 0 | 1 | 0.212 | 0.409 |
| A33 | 0 | 1 | 0.782 | 0.413 |
| A34 | 0 | 1 | 0.624 | 0.485 |
| A35 | 0 | 1 | 0.339 | 0.473 |
| A36 | 0 | 1 | 0.234 | 0.423 |
Before applying IRT, we should check that the data satisfies the assumptions needed by the model. In particular, to use a 1-dimensional IRT model, we should have some evidence of unidimensionality in the test scores.
This plot shows the correlations between scores on each pair of items:
item_scores <- test_scores %>%
select(-class, -year)
cor_ci <- psych::corCi(item_scores, plot = FALSE)
psych::cor.plot.upperLowerCi(cor_ci)
There are a few correlations that are not significantly different from 0:
cor_ci$ci %>%
as_tibble(rownames = "corr") %>%
filter(p > 0.05) %>%
arrange(-p) %>%
select(-contains(".e")) %>%
gt() %>%
fmt_number(columns = 2:4, decimals = 3)
| corr | lower | upper | p |
|---|---|---|---|
| A2-A34 | −0.034 | 0.058 | 0.615 |
| A1-A2 | −0.021 | 0.056 | 0.366 |
| A1-A36 | −0.013 | 0.069 | 0.186 |
| A1-A18 | −0.003 | 0.079 | 0.070 |
The overall picture is that the item scores are well correlated with each other.
structure <- check_factorstructure(item_scores)
n <- n_factors(item_scores)
The choice of 1 dimensions is supported by 4 (17.39%) methods out of 23 (Acceleration factor, VSS complexity 1, Velicer’s MAP, RMSEA).
plot(n)
summary(n) %>% gt()
| n_Factors | n_Methods |
|---|---|
| 1 | 4 |
| 2 | 1 |
| 3 | 1 |
| 4 | 1 |
| 5 | 2 |
| 6 | 4 |
| 9 | 1 |
| 11 | 2 |
| 26 | 1 |
| 28 | 1 |
| 33 | 1 |
| 34 | 1 |
| 35 | 3 |
#n %>% tibble() %>% gt()
fa.parallel(item_scores, fa = "fa")
## Parallel analysis suggests that the number of factors = 9 and the number of components = NA
fitfact <- factanal(item_scores, factors = 1, rotation = "varimax")
print(fitfact, digits = 2, cutoff = 0.3, sort = TRUE)
##
## Call:
## factanal(x = item_scores, factors = 1, rotation = "varimax")
##
## Uniquenesses:
## A1 A2 A3 A4 A5 A6 A7 A8 A9 A10 A11 A12 A13 A14 A15 A16
## 0.95 0.96 0.76 0.82 0.74 0.76 0.82 0.80 0.63 0.69 0.68 0.90 0.92 0.76 0.88 0.81
## A17 A18 A19 A20 A21 A22 A23 A24 A25 A26 A27 A28 A29 A30 A31 A32
## 0.76 0.81 0.80 0.64 0.61 0.66 0.66 0.75 0.93 0.70 0.64 0.68 0.88 0.83 0.80 0.82
## A33 A34 A35 A36
## 0.87 0.88 0.81 0.86
##
## Loadings:
## [1] 0.51 0.61 0.56 0.56 0.60 0.62 0.58 0.58 0.55 0.60 0.57 0.49 0.43
## [16] 0.49 0.42 0.44 0.31 0.49 0.35 0.43 0.49 0.44 0.45 0.50 0.35 0.41
## [31] 0.45 0.43 0.36 0.35 0.44 0.38
##
## Factor1
## SS loadings 7.75
## Proportion Var 0.22
##
## Test of the hypothesis that 1 factor is sufficient.
## The chi square statistic is 3903.11 on 594 degrees of freedom.
## The p-value is 0
load <- tidy(fitfact)
ggplot(load, aes(x = fl1, y = 0)) +
geom_point() +
geom_label_repel(aes(label = paste0("A", rownames(load))), show.legend = FALSE) +
labs(x = "Factor 1", y = NULL,
title = "Standardised Loadings",
subtitle = "Based upon correlation matrix") +
theme_minimal()
## Warning: ggrepel: 14 unlabeled data points (too many overlaps). Consider
## increasing max.overlaps
# To proceed with the IRT analysis, comment out the following line before knitting
#knitr::knit_exit()
We can fit a Multidimensional Item Response Theory (mirt) model. From the function definition:
mirt fits a maximum likelihood (or maximum a posteriori) factor analysis model to any mixture of dichotomous and polytomous data under the item response theory paradigm using either Cai's (2010) Metropolis-Hastings Robbins-Monro (MHRM) algorithm.
The process is to first fit the model, and save the result as a model object that we can then parse to get tabular or visual displays of the model that we might want. When fitting the model, we have the option to specify a few arguments, which then get interpreted as parameters to be passed to the model.
fit_2pl <- mirt(
data = item_scores, # just the columns with question scores
model = 1, # number of factors to extract
itemtype = "2PL", # 2 parameter logistic model
SE = TRUE # estimate standard errors
)
##
Iteration: 1, Log-Lik: -42537.007, Max-Change: 0.80697
Iteration: 2, Log-Lik: -41857.079, Max-Change: 0.42644
Iteration: 3, Log-Lik: -41765.562, Max-Change: 0.39313
Iteration: 4, Log-Lik: -41733.247, Max-Change: 0.13739
Iteration: 5, Log-Lik: -41714.377, Max-Change: 0.08694
Iteration: 6, Log-Lik: -41703.599, Max-Change: 0.06036
Iteration: 7, Log-Lik: -41697.345, Max-Change: 0.05323
Iteration: 8, Log-Lik: -41693.703, Max-Change: 0.03370
Iteration: 9, Log-Lik: -41691.469, Max-Change: 0.03145
Iteration: 10, Log-Lik: -41689.421, Max-Change: 0.01847
Iteration: 11, Log-Lik: -41688.919, Max-Change: 0.01498
Iteration: 12, Log-Lik: -41688.609, Max-Change: 0.01278
Iteration: 13, Log-Lik: -41688.333, Max-Change: 0.00703
Iteration: 14, Log-Lik: -41688.246, Max-Change: 0.00468
Iteration: 15, Log-Lik: -41688.188, Max-Change: 0.00459
Iteration: 16, Log-Lik: -41688.115, Max-Change: 0.00298
Iteration: 17, Log-Lik: -41688.094, Max-Change: 0.00173
Iteration: 18, Log-Lik: -41688.087, Max-Change: 0.00123
Iteration: 19, Log-Lik: -41688.075, Max-Change: 0.00113
Iteration: 20, Log-Lik: -41688.071, Max-Change: 0.00065
Iteration: 21, Log-Lik: -41688.068, Max-Change: 0.00077
Iteration: 22, Log-Lik: -41688.064, Max-Change: 0.00052
Iteration: 23, Log-Lik: -41688.062, Max-Change: 0.00041
Iteration: 24, Log-Lik: -41688.061, Max-Change: 0.00023
Iteration: 25, Log-Lik: -41688.060, Max-Change: 0.00025
Iteration: 26, Log-Lik: -41688.059, Max-Change: 0.00024
Iteration: 27, Log-Lik: -41688.058, Max-Change: 0.00022
Iteration: 28, Log-Lik: -41688.056, Max-Change: 0.00016
Iteration: 29, Log-Lik: -41688.055, Max-Change: 0.00016
Iteration: 30, Log-Lik: -41688.055, Max-Change: 0.00015
Iteration: 31, Log-Lik: -41688.054, Max-Change: 0.00012
Iteration: 32, Log-Lik: -41688.054, Max-Change: 0.00012
Iteration: 33, Log-Lik: -41688.054, Max-Change: 0.00012
Iteration: 34, Log-Lik: -41688.053, Max-Change: 0.00010
##
## Calculating information matrix...
We then compute factor score estimates and augment the existing data frame with these estimates, to keep everything in one place. To do the estimation, we use the fscores() function from the mirt package which takes in a computed model object and computes factor score estimates according to the method specified. We will use the EAP method for factor score estimation, which is the “expected a-posteriori” method, the default. We specify it explicitly below, but the results would have been the same if we omitted specifying the method argument since it’s the default method the function uses.
test_scores <- test_scores %>%
mutate(F1 = fscores(fit_2pl, method = "EAP"))
We can also calculate the model coefficient estimates using a generic function coef() which is used to extract model coefficients from objects returned by modeling functions. We will set the IRTpars argument to TRUE, which means slope intercept parameters will be converted into traditional IRT parameters.
coefs_2pl <- coef(fit_2pl, IRTpars = TRUE)
The resulting object coefs is a list, with one element for each question, and an additional GroupPars element that we won’t be using. The output is a bit long, so we’re only showing a few of the elements here:
coefs_2pl[1:3]
## $A1
## a b g u
## par 1.277290 -3.136814 0 1
## CI_2.5 1.003766 -3.629126 NA NA
## CI_97.5 1.550813 -2.644502 NA NA
##
## $A2
## a b g u
## par 0.4686183 1.969524 0 1
## CI_2.5 0.3634940 1.517563 NA NA
## CI_97.5 0.5737426 2.421485 NA NA
##
## $A3
## a b g u
## par 1.320866 -0.6360125 0 1
## CI_2.5 1.172386 -0.7343667 NA NA
## CI_97.5 1.469346 -0.5376583 NA NA
# coefs_2pl[35:37]
Let’s take a closer look at the first element:
coefs_2pl[1]
## $A1
## a b g u
## par 1.277290 -3.136814 0 1
## CI_2.5 1.003766 -3.629126 NA NA
## CI_97.5 1.550813 -2.644502 NA NA
In this output:
a is discriminationb is difficultyTo make this output a little more user friendly, we should tidy it such that we have a row per question. We’ll do this in two steps. First, write a function that tidies the output for one question, i.e. one list element. Then, map this function over the list of all questions, resulting in a data frame.
tidy_mirt_coefs <- function(x){
x %>%
# melt the list element
melt() %>%
# convert to a tibble
as_tibble() %>%
# convert factors to characters
mutate(across(where(is.factor), as.character)) %>%
# only focus on rows where X2 is a or b (discrimination or difficulty)
filter(X2 %in% c("a", "b")) %>%
# in X1, relabel par (parameter) as est (estimate)
mutate(X1 = if_else(X1 == "par", "est", X1)) %>%
# unite columns X2 and X1 into a new column called var separated by _
unite(X2, X1, col = "var", sep = "_") %>%
# turn into a wider data frame
pivot_wider(names_from = var, values_from = value)
}
Let’s see what this does to a single element in coefs:
tidy_mirt_coefs(coefs_2pl[1])
## # A tibble: 1 x 7
## L1 a_est a_CI_2.5 a_CI_97.5 b_est b_CI_2.5 b_CI_97.5
## <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 A1 1.28 1.00 1.55 -3.14 -3.63 -2.64
And now let’s map it over all 32 elements of coefs:
tidy_2pl <- map_dfr(coefs_2pl[1:32], tidy_mirt_coefs, .id = "Question")
A quick peek at the result:
tidy_2pl
## # A tibble: 32 x 7
## Question a_est a_CI_2.5 a_CI_97.5 b_est b_CI_2.5 b_CI_97.5
## <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 A1 1.28 1.00 1.55 -3.14 -3.63 -2.64
## 2 A2 0.469 0.363 0.574 1.97 1.52 2.42
## 3 A3 1.32 1.17 1.47 -0.636 -0.734 -0.538
## 4 A4 1.09 0.955 1.22 -0.671 -0.785 -0.557
## 5 A5 1.41 1.26 1.56 -0.00621 -0.0875 0.0751
## 6 A6 1.50 1.33 1.67 -0.989 -1.10 -0.881
## 7 A7 1.15 1.01 1.29 -1.08 -1.22 -0.949
## 8 A8 1.11 0.979 1.24 -0.0839 -0.178 0.0106
## 9 A9 1.97 1.77 2.17 0.0409 -0.0279 0.110
## 10 A10 1.64 1.47 1.81 -0.200 -0.276 -0.124
## # ... with 22 more rows
And a nicely formatted table of the result:
gt(tidy_2pl) %>%
fmt_number(columns = contains("_"), decimals = 3) %>%
data_color(
columns = contains("a_"),
colors = scales::col_numeric(palette = c("Greens"), domain = NULL)
) %>%
data_color(
columns = contains("b_"),
colors = scales::col_numeric(palette = c("Blues"), domain = NULL)
) %>%
tab_spanner(label = "Discrimination", columns = contains("a_")) %>%
tab_spanner(label = "Difficulty", columns = contains("b_")) %>%
cols_label(
a_est = "Est.",
b_est = "Est.",
a_CI_2.5 = "2.5%",
b_CI_2.5 = "2.5%",
a_CI_97.5 = "97.5%",
b_CI_97.5 = "97.5%"
)
tidy_2pl %>%
mutate(qnum = parse_number(Question)) %>%
ggplot(aes(x = qnum, y = b_est)) +
geom_errorbar(aes(ymin = b_CI_2.5, ymax = b_CI_97.5), width = 0.2) +
geom_point() +
theme_minimal() +
labs(x = "Question",
y = "Difficulty")
Do students from different programmes of study have different distributions of ability?
Compare the distribution of abilities in the year groups (though in this case there is only one).
ggplot(test_scores, aes(F1, fill = as.factor(year), colour = as.factor(year))) +
geom_density(alpha=0.5) +
scale_x_continuous(limits = c(-3.5,3.5)) +
labs(title = "Density plot",
subtitle = "Ability grouped by year of taking the test",
x = "Ability", y = "Density",
fill = "Year", colour = "Year") +
theme_minimal()
Compare the distribution of abilities in the various classes.
ggplot(test_scores, aes(x = F1, y = class, colour = class, fill = class)) +
geom_density_ridges(alpha = 0.5) +
scale_x_continuous(limits = c(-3.5,3.5)) +
guides(fill = FALSE, colour = FALSE) +
labs(title = "Density plot",
subtitle = "Ability grouped by class of taking the test",
x = "Ability", y = "Class") +
theme_minimal()
## Picking joint bandwidth of 0.286
plot(fit_2pl, type = "infoSE", main = "Test information")
plot(fit_2pl, type = "infotrace", main = "Item information curves")
plot(fit_2pl, type = "score", auto.key = FALSE)
We can get individual item surface and information plots using the itemplot() function from the mirt package, e.g.
mirt::itemplot(fit_2pl, item = 1,
main = "Trace lines for item 1")
We can also get the plots for all trace lines, one facet per plot.
plot(fit_2pl, type = "trace", auto.key = FALSE)
Or all of them overlaid in one plot.
plot(fit_2pl, type = "trace", facet_items=FALSE)
An alternative approach is using ggplot2 and plotly to add interactivity to make it easier to identify items.
# store the object
plt <- plot(fit_2pl, type = "trace", facet_items = FALSE)
# the data we need is in panel.args
# TODO - I had to change the [[1]] to [[2]] since the plt has two panels for some reason, with the one we want being the 2nd panel
plt_data <- tibble(
x = plt$panel.args[[2]]$x,
y = plt$panel.args[[2]]$y,
subscripts = plt$panel.args[[2]]$subscripts,
item = rep(colnames(item_scores), each = 200)
) %>%
mutate(
item_no = str_remove(item, "A") %>% as.numeric(),
item = fct_reorder(item, item_no)
)
head(plt_data)
## # A tibble: 6 x 5
## x y subscripts item item_no
## <dbl> <dbl> <int> <fct> <dbl>
## 1 -6 0.0252 201 A1 1
## 2 -5.94 0.0271 202 A1 1
## 3 -5.88 0.0292 203 A1 1
## 4 -5.82 0.0315 204 A1 1
## 5 -5.76 0.0339 205 A1 1
## 6 -5.70 0.0365 206 A1 1
plt_gg <- ggplot(plt_data, aes(x, y,
colour = item,
text = item)) +
geom_line() +
labs(
title = "2PL - Trace lines",
#x = expression(theta),
x = "theta",
#y = expression(P(theta)),
y = "P(theta)",
colour = "Item"
) +
theme_minimal() +
theme(legend.position = "none")
ggplotly(plt_gg, tooltip = "text")
knitr::knit_exit()